Consider the data for the following reaction at 25 o C 2 HgCl 2 (aq) + C 2 O 4 2
ID: 876275 • Letter: C
Question
Consider the data for the following reaction at 25 oC
2 HgCl2(aq) + C2O42-(aq) --------2 Cl-(aq) + 2 CO2 (g) + Hg2Cl2(s) Where Rate = [C2O42-]/t
Determine the 'orders for each reactant.
Determine the rate constant for this reaction.
Determine the instantaneous rate of change for HgCl2 when [HgCl2] = O.l22 M and [C2O42- ] = 0.50 M.
Experiment
[HgCl2] (M)
[C2O42-] (M)
Rate (M/s)
1
0.164
0.15
3.2 x 10-5
2
0.164
0.45
2.9 x 10-5
3
0.082
0.45
1.4 x 10-5
4
0.246
0.15
4.8 x 10-5
Experiment
[HgCl2] (M)
[C2O42-] (M)
Rate (M/s)
1
0.164
0.15
3.2 x 10-5
2
0.164
0.45
2.9 x 10-5
3
0.082
0.45
1.4 x 10-5
4
0.246
0.15
4.8 x 10-5
Explanation / Answer
let the rate law be
rate = k [HgCl2]^a [C2042-]^b
now
consider experiment 1 and 2
we see that
[HgCl2] is constant
k is already a constant
so
rate2/rate1 = ( [C2042-]2/ [C2042-]1)^b
2.9 x 10-5 / 3.2 x 10-5 = ( 0.45 / 0.15)^b
b = 0
the order with respect to C2042- is 0
now
consider experiment 2 and 3
we see that
[C2042-] is constant
k is already a constant
so
rate 3 / rate 2 = ( [HgCl2]3/[HgCl2]2)^a
so
1.4 x 10-5 / 2.9 x 10-5 = ( 0.082/0.164)^a
solving we get
a = 1
the order with respect to HgCl2 is 1
so
the rate law is
rate = k [HgCl2]^1 [C2042-]^0
rate = k [HgCl2]
now
conisder experiment 1
we get
3.2 x 10-5 = k [0.164]
so
k= 1.95 x 10-4
so
the rate constant is 1.95 x 10-4
now
given [HgCl2] = 0.122 M
so
rate = k [HgCl2]
rate = 1.95 x 10-4 x 0.122
rate = 2.38 x 10-5
so
the rate when [HgCL2] =0.122 M is 2.38 x 10-5